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Math Help - convergent sequences

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    Last edited by britnumber; October 9th 2008 at 08:34 PM.
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    Quote Originally Posted by britnumber View Post
    Let X=(x_n) be a sequence of positive real numbers such that = L >1. Show that X is not a bounded sequence and hence is not convergent
    Let \epsilon > 0 so that L - \epsilon > 1.

    And therefore there is N so that if n\geq N then \left| \frac{x_{n+1}}{x_n} - L \right| < \epsilon.

    Therefore, L - \epsilon< \frac{x_{n+1}}{x_n} \implies (L - \epsilon)x_n < x_{n+1}.

    Therefore,
    (L - \epsilon)x_N < x_{N+1}
    (L - \epsilon)x_{N+1} < x_{N+2}
    (L - \epsilon)x_{N+2} < x_{N+3}
    ...
    Thus,
    (L - \epsilon)^2x_N <  x_{N+2}
    (L - \epsilon)^3x_N < x_{N+3}
    ...
    In general,
    (L - \epsilon)^kx_N < x_{N+k}

    But (L - \epsilon)^k can be made arbitary large since L - \epsilon > 1.
    Therefore, \{ x_n\} is unbounded.
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